What is the remainder when a positive integer P is divided by 20?

(1) P = X^2 where X is a positive integer
X=2, P=4, remainder is 4
X=4, P=16, remainder is 16
NOT SUFFICIENT

(2) X is the square of an even integer
We have to know P, not X
NOT SUFFICIENT

(1)+(2)
X=4, P=16, remainder is 16
X=16, P=256, remainder is 16
X=36, P=1296, remainder is 16
SUFFICIENT

FINAL ANSWER IS (C)

What is the remainder when a positive integer P is divided by 20?

(1) P = X^2 where X is a positive integer
(2) X is the square of an even integer


i) \(P=X^2\) => P can be 1,4,16,25 with number of remainders - NS
ii) Nothing about P - NS

Combining i) and ii)
X = \((2m)^2\); P = \(X^2\) => \(P=16*m^4\) (where m can be any positive integer)

Therefore, \(\frac{P}{20}\) will give remainder of 16 or (-4) from the first term.

Let's check for \(m^4\),
m = 1 => remainder 1 => \(\frac{P}{20}\) will give remainder of -4 = 16
m = 2 => remainder \(2^4\) = 16 = -4 => \(\frac{P}{20}\) will give remainder of -4*-4 = 16
m = 5 => remainder of \(\frac{625}{20}\) = 5 => \(\frac{P}{20}\) will give remainder of -4*5 = -20 = 0

2 different answers - Not Sufficient

Answer E

What is the remainder when a positive integer P is divided by 20?

(1) P = X^2 where X is a positive integer
(2) X is the square of an even integer

From statement (1), let x = 1, 2, 3, 4, 5, 6, …………
P = x^2 = 1, 4, 9, 25, 36, …………………
When p/25, then remainder = 1, 4 , 5 , 16 …………….. Not sufficient.

From statement (2), there is no P. Hence, Not Sufficient.

Combining both statements, we get, P = x^2 = 4, 16, 36, 84, 100, 144, ………..
When p/25, Remainder = 4, 16, 16, 4, 0, 4, Not Sufficient.
Answer: E

What is the remainder when a positive integer P is divided by 20?

(Statement1): \(P = X^2\) where X is a positive integer
--> \(P=6^{2}\) --> \(\frac{36}{20}= 1\) (remainder-16)
--> \(P=10^{2}\)--> \(\frac{100}{20}=5\) (remainder-0)
Insufficient

(Statement2): X is the square of an even integer
the question is about P, not X. Nothing tells us about what P is.
Insufficient

Taken together 1&2,
\(P= (4^{2})^{2} =256\) --> \(\frac{256}{20}= 12\) (remainder -16)
\(P= (10^{2})^{2}= 10000\) --> \(\frac{10000}{20} =500\) (remainder -0)
Still we got more than one value.
Insufficient

The answer is E.

We are given that P is a positive integer, and we are to determine the remainder when P is divided by 20.

Statement 1: P=X^2, where X is a positive integer.
Possible values of P: 1, 4, 9, 16, 25, 36, etc.
When P=1, the remainder is 1. When P=4, the remainder is 4.
Statement 1 is insufficient since we have more than one possibility as the remainder when P is divided by 20.

Statement 2: X is the square of an even integer.
X=4, 16, 36, 64, 100, etc.
Clearly statement 2 is insufficient because we have no idea the value of P, and as per the question, there are many possibilities that lead to 20 possible remainders when P is divided by 20.

1+2
From 1, we know that P=X^2 and from 2 we know that X can be: 4, 16, 36, 64, 100, etc
Hence P=16, 256, or 10,000, etc.

When P=16, the remainder is 16, however, when P=10,000, the remainder is 0.
Both statements even when taken together are not sufficient.

E is the answer.

Hi Bunuel,

Can you explain why it is E? Statement 2 says X is the square of an even integer, so X=2 must not be included. Thus, X=4,16,36,64,... and P=16,256,1296,4096,... Whenever we divide P, the remainder is always 16. Thus, answer should be C.

Thanks for clarification

P is the fourth power of a positive even integer.

If P = 2^4, then the remainder is 16.
If P = 20^4, then the remainder is 0.

remainder: p/20

(1) p=x^2, x=pos integer; insufic.
p=1,4,9… remainder=1,4,9…

(2) x=4,16,36,64,100… p=? insufic

(1/2) insufic
p=x^2=(4,16,36,64,100…)^2
p=4^2=16; remainder=16
p=100^2; remainder=0

Ans (E)

This approach is wrong because are you going to try every single number from 4 to 8 until you reached 10? multiplying all those numbers would have taken well over 2 mins. You have to realize that 10 to the 2nd power is 100. and that to the next power will be divisible by 20.

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